Question

If $A$ and $B$ are positive acute angles satisfying $3{\cos }^{2}A+2{\cos }^{2}B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$.
Then the value of $A+2B$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

Given, $3{\cos }^{2}A+2{\cos }^{2}B=4$
$\Rightarrow 2{\cos }^{2}B-1=4-3{\cos }^{2}A-1$
$\Rightarrow \cos 2B=3(1-{\cos }^{2}A)=3{\sin }^{2}A$ ..... (1)
and $2\cos B\sin B=3\sin A\cos A$
$\sin 2B=3\sin A\cos A$ .... (2)
Now, $\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B$
$=\cos A\left(3{\sin }^{2}A\right)-\sin A\left(3\sin A\cos A\right)=0$ ....[using eqs. (1) and (2)]
$\Rightarrow A+2B=\frac{\pi }{2}$